Problems

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Found: 11

Two different numbers \(x\) and \(y\) (not necessarily integers) are such that \(x^2-2000x=y^2-2000y\). Find the sum of \(x\) and \(y\).

To each pair of numbers \(x\) and \(y\) some number \(x * y\) is placed in correspondence. Find \(1993 * 1935\) if it is known that for any three numbers \(x, y, z\), the following identities hold: \(x * x = 0\) and \(x * (y * z) = (x * y) + z\).

Prove that for any natural number \(a_1> 1\) there exists an increasing sequence of natural numbers \(a_1, a_2, a_3, \dots\), for which \(a_1^2+ a_2^2 +\dots+ a_k^2\) is divisible by \(a_1+ a_2+\dots+ a_k\) for all \(k \geq 1\).

A numeric set \(M\) containing 2003 distinct numbers is such that for every two distinct elements \(a, b\) in \(M\), the number \(a^2+ b\sqrt 2\) is rational. Prove that for any \(a\) in \(M\) the number \(q\sqrt 2\) is rational.

Prove that, if \(b=a-1\), then \[(a+b)(a^2 +b^2)(a^4 +b^4)\dotsb(a^{32} +b^{32})=a^{64} -b^{64}.\]

Prove the following formulae are true: \[\begin{aligned} a^{n + 1} - b^{n + 1} &= (a - b) (a^n + a^{n-1}b + \dots + b^n);\\ a^{2n + 1} + b^{2n + 1} &= (a + b) (a^{2n} - a^{2n-1}b + a^{2n-2}b^2 - \dots + b^{2n}). \end{aligned}\]

The number \(x\) is such a number that exactly one of the four numbers \(a = x - \sqrt{2}\), \(b = x-1/x\), \(c = x + 1/x\), \(d = x^2 + 2\sqrt{2}\) is not an integer. Find all such \(x\).